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Question: Two triodes having amplification factor 30 and 21 and plate resistance \(5k\Omega \) and \(4k\Omega ...

Two triodes having amplification factor 30 and 21 and plate resistance 5kΩ5k\Omega and 4kΩ4k\Omega respectively are connected in parallel. The composite amplification factor of the system is
A.25 B.50 C.75 D.100 \begin{aligned} & A.25 \\\ & B.50 \\\ & C.75 \\\ & D.100 \\\ \end{aligned}

Explanation

Solution

Amplification factor of a triode is nothing but the factor by which an input signal is enhanced. For a triode, amplification factor is the rate of change of anode current with respect to grid voltage. It can be calculated as the product of transconductance and plate resistance. The effective transconductance of two triodes connected in parallel is simply the sum of individual transconductance. The effective plate resistance of two triodes connected in parallel can be calculated by the formula rP=rP1rP2rP1+rP2{{r}_{P}}=\dfrac{{{r}_{{{P}_{1}}}}{{r}_{{{P}_{2}}}}}{{{r}_{{{P}_{1}}}}+{{r}_{{{P}_{2}}}}}where, rP1{{r}_{{{P}_{1}}}} and rP2{{r}_{{{P}_{2}}}} are the individual plate resistances.

Complete answer:
A triode is an electronic device used for amplification purpose. It consists of a hot filament which acts as a cathode and it releases electrons when it is heated sufficiently. It also has a metal plate electrode which acts as anode and attracts the electrons emitted by cathode.
If μ\mu is the amplification factor of a triode, rP{{r}_{P}}is the plate resistance then
μ=rPgm\mu ={{r}_{P}}{{g}_{m}}
Where gm{{g}_{m}} is the transconductance of the triode.
Therefore,
gm=μrP{{g}_{m}}=\dfrac{\mu }{{{r}_{P}}}
Given that
μ1=30 rP1=5kΩ=5×10−3Ω \begin{aligned} & {{\mu }_{1}}=30 \\\ & {{r}_{{{P}_{1}}}}=5k\Omega =5\times {{10}^{-3}}\Omega \\\ \end{aligned}
μ2=21 rP2=4kΩ=4×10−3Ω \begin{aligned} & {{\mu }_{2}}=21 \\\ & {{r}_{{{P}_{2}}}}=4k\Omega =4\times {{10}^{-3}}\Omega \\\ \end{aligned}
Substituting this values in above equation, we get,
gm1=μ1rP1=305×10−3=6×10−3mho{{g}_{{{m}_{1}}}}=\dfrac{{{\mu }_{1}}}{{{r}_{{{P}_{1}}}}}=\dfrac{30}{5\times {{10}^{-3}}}=6\times {{10}^{-3}}mho
and
gm2=μ2rP2=214×10−3=5.25×10−3mho{{g}_{{{m}_{2}}}}=\dfrac{{{\mu }_{2}}}{{{r}_{{{P}_{2}}}}}=\dfrac{21}{4\times {{10}^{-3}}}=5.25\times {{10}^{-3}}mho
Therefore, effective transconductance is
gm=gm1+gm2 gm=(6+5.25)×10−3mho gm=11.25×10−3mho \begin{aligned} & {{g}_{m}}={{g}_{{{m}_{1}}}}+{{g}_{{{m}_{2}}}} \\\ & {{g}_{m}}=(6+5.25)\times {{10}^{-3}}mho \\\ & {{g}_{m}}=11.25\times {{10}^{-3}}mho \\\ \end{aligned}
Since triodes are connected in parallel, there effective resistance is given as
rP=rP1rP2rP1+rP2 rP=5×45+4kΩ rP=209kΩ rP=209×10−3Ω \begin{aligned} & {{r}_{P}}=\dfrac{{{r}_{{{P}_{1}}}}{{r}_{{{P}_{2}}}}}{{{r}_{{{P}_{1}}}}+{{r}_{{{P}_{2}}}}} \\\ & {{r}_{P}}=\dfrac{5\times 4}{5+4}k\Omega \\\ & {{r}_{P}}=\dfrac{20}{9}k\Omega \\\ & {{r}_{P}}=\dfrac{20}{9}\times {{10}^{-3}}\Omega \\\ \end{aligned}
Therefore, the composite amplification factor is given by
μ=rPgm μ=11.25×10−3×209×10−3 μ=25 \begin{aligned} & \mu ={{r}_{P}}{{g}_{m}} \\\ & \mu =11.25\times {{10}^{-3}}\times \dfrac{20}{9}\times {{10}^{-3}} \\\ & \mu =25 \\\ \end{aligned}
Thus, the composite amplification factor is 25.
Answer - A. 25

Note: The amplification factor of a triode is independent of the voltage at its anode and cathode. It depends on its geometric conditions. It is the product of transconductance and its plate resistance. The value of amplification factor generally ranges from 10 to 100. The transconductance of grid plates is also termed as mutual conductance. Note that amplification factor is a unitless and dimensionless quantity.